wer:Given that,mass = m = 20 kginitial velocity = u = 40 m/sFinal velocity = v = 50 m/sTime taken = t = 2 sWe know that,a = (v - u)/tPutting the values given in the question,a = (50 - 40)/2→ a = 10/2→ a = 5 m/s²We know that,s = ut + (1/2)at²Putting the values given in the question,→ s = (40 * 2) + (1/2)(5 * 2²)→ s = 80 + 10→ s = 90 MWE know that,F = maPutting the values given in the question,→ F = 20 * 5→ F = 100 NWe know that,W = F s cos∅where W is WORK done, F is force, s is displacement and ∅ is the angle between force and displacement. Considering the force and displacement are in same direction, ∅ = 0°.→ W = 100 * 90 * cos0°→ W = 9000 J = 9 kJ ANS.2nd answer:Given that,mass = m = 35 kginitial height = h₁ = 4 mFinal height = h₂ = 12 mWe can consider any point to be a reference point and SAY that the gravitational potential energy (GPE) at that point will be 0. The GPE at any other point will be calculated in reference to the reference point. We know that,GPE = mghwhere m is the mass, g is the acceleration due to gravity, and h is the displacement between where the object is and the reference point. Considering the ground level to be reference point,Initial GPE = mgh₁Final GPE = mgh₂Then, change in GPE = mgh₂ - mgh₁Putting the values given in the question,→ ∆GPE = mg(h₂ - h₁)→ ∆GPE = 35 * 10 * (12 - 4)→ ∆GPE = 35 * 10 * 8→ ∆GPE = 2800 J = 2.8 kJ Ans.